Colby College
Gibbs Phase Rule: f = c – p + 2
f = Intensive Degrees of freedom = variance
Number of intensive variables that can be changed independently without
disturbing the number of phases in equilibrium
p = number of phases
gas, homogeneous liquid phases, homogeneous solid phases
c = components
Minimum number of independent constituents
Case I. No chemical reactions: c = constituents
Example 1: start with methanol and water – 2 components
Case II With chemical reactions:
Example 2: start with NaH
2
PO
4
in water --
K
a2
K
a3
H
2
PO
4-
→
←
HPO
42-
+ H
+
→
←
PO
43-
+ H
+
Constituents: Na
+
, H
+
, H
2
PO
4-
, HPO
42-
, PO
43-
, H
2
O
but only 2 components -- NaH
2
PO
4
and H
2
O.
Example 3: start with NaH
2
PO
4
and Na
2
HPO
4
in water --
Same constituents: Na
+
, H
+
, H
2
PO
4-
, HPO
42-
, PO
43-
, H
2
O
but now 3 components -- NaH
2
PO
4
, Na
2
HPO
4
, and H
2
O.
Need to know: T, P, y
A
, y
B
, x
A
, x
B
total intensive variables = c p + 2
But y
A
+ y
B
= 1
x
A
+ x
B
= 1
Get p such equations, one for each phase:
Independent variables = c p + 2 – p
But, chemical potential is everywhere equal:
µ
A
(x
A
) = µ
A
(g)
µ
B
(x
B
) = µ
B
(g)
Get p–1 for each component
Get c( p–1) such equations:
Independent variables = c p + 2 – p – c( p–1)
f = c – p + 2
A & B
liquid
µ
A
(g)
µ
A
(x
A
)
=
µ
B
(x
B
)
=
µ
B
(g)
x
A
+ x
B
=1
A & B
vapor
y
A
+ y
B
=1
f ' = c – p + 1
P = cst
0 1
x
A
,
y
A
→
liquid A & B
f ' = 2
vapor A & B
T f ' = 1
f ' = 2
T
*
bA
T
*
bB
